4 Answers
4

The general element is $\pm\exp(A)$ where $A$ is skew-symmetric.
(This gives each element infinitely often). This trick essentially
works for all compact Lie groups.

There is also the Cayley parameterization: $(I+A)(I-A)^{-1}$
for skew-symmetric $A$
is the general element of $SO(3)$ which lacks an eigenvalue $-1$
(so isn't a half-turn.) This parameterizes all such matrices
once each.

First, if $A\in O(3)$ then either $A$ or $-A$ is in $SO(3)$ so you can just restrict attention to $SO(3)$. For that, one answer is the Cayley transform: let $so(3)$ denote the set of $3\times 3$ antisymmetric matrices (which is homeomorphic to $\mathbb{R}^3$) and put

$U=\{R\in SO(3) : \det(R+I)\neq 0\}$

Then $U$ is a dense open subset of $SO(3)$ and there is a homeomorphism $f:so(3)\to U$ given by $f(A)=(I+A)(I-A)^{-1}$.

I assume that you are interested in the case of matrices with real entries. Perhaps one should reduce the problem to $SO(3)$, since $O(3)$ has two connected components, or we may use the determinant $\pm1$ as one parameter. A natural parametrization is unlikely, or will not be injective, because $SO(3)$ has a non-trivial topology: its fundamental group is $\mathbb Z/2\mathbb Z$. One possibility is to give a direction $\vec u$ (a unitary vector) of the axis of rotation, and the angle $\theta\in\mathbb R/2\pi\mathbb Z$ of this rotation. But then$(-\vec u,\theta+\pi)\sim(\vec u,\theta)$, and all pairs $(\vec u,0)$ yield the same element $I_3$.

If there were a really simple way we wouldn't need the concept of "gimbal lock" (http://en.wikipedia.org/wiki/Gimbal_lock). In other words the manifold in question is compact but isn't the 3-torus, and pretending it is has to break down somewhere. The unit quaternions are the "slickest" surrogate, but "parametrize" implies charts and global charts aren't going to work out "simply".